Four charges of equal magnitude Q = 55 nC are placed on the corners of a rectang

Question

Four charges of equal magnitude Q = 55 nC are placed on the corners of a rectangle of sides D_1 = 28 cm and D_2 = 12 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. For this problem, use the coordinate system and angle indicated. Which of the following represents a free-body diagram for the charge on the lower left hand corner of the rectangle? Write an equation for the net horizontal force acting on the charge located at the lower left comer of the rectangle in terms of the magnitude of the forces from the three other charges, F_1, F_2, F_3, and the angle theta. Calculate the numeric value for the horizontal force, in newtons. Write an equation for the net vertical force acting on the charge located at the lower left comer of the rectangle in terms of the magnitude of the forces from the three other charges, F_1. F_2, F_3, and the angle theta. Calculate the numeric value for the vertical force, in newtons. Calculate the magnitude of the net force on the charge located at the lower left comer of the rectangle, in newtons. Calculate the angle, in degrees counterclockwise with respect to the positive x-axis, of the net force. theta

Explanation / Answer

part c:

angle theta=arctan(D2/D1)=arctan(12/28)=23.2 degrees

force on lower left charge due to upper left charge:

as both charges are of same sign, force is repulsive in nature and will act in -ve y axis direction.

force magnitude=F1=k*Q*Q/D2^2

where k=coloumb's constant

F1=9*10^9*55*10^(-9)*55*10^(-9)/0.12^2=1.8906*10^(-3) N

in vector notation, force F1=-(1.8906*10^(-3) j ) N

force on lower left charge due to lower right charge:

as charges are of opposite sign, force is attractive in nature.

and directed towards +ve x axis.

force magnitude=F2=k*Q*Q/D1^2

=9*10^9*55*10^(-9)*55*10^(-9)/0.28^2

=3.4726*10^(-4) N

in vector notation force F2=(3.4726*10^(-4) i ) N

force on lower left charge due to upper right charge:

as charges are of opposite sign, force is attractive in nature and directed towards the -ve charge.

force direction in vector notation=cos(theta) i + sin (theta) j

=0.9191 i + 0.39394 j

force magnitude=F3=k*Q*Q/(D1^2+D2^2)

=9*10^9*55*10^(-9)*55*10^(-9)/(0.12^2+0.28^2)

=2.9337*10^(-4) N

in vector notation , force F3=2.9337*10^(-4)*(0.9191 i + 0.39394 j)

=2.69636*10^(-4) i + 1.1557*10^(-4) j

total force=-F1 j + F2 i + F3*(cos(theta) i + sin(theta) j)

horizontal force=F2+F3*cos(theta)

=3.4726*10^(-4)+2.69636*10^(-4)=6.169*10^(-4) N

part d:

vertical force=-F1+F3*sin(theta)

part e:

value of vertical force=-1.8906*10^(-3)+1.1557*10^(-4)=-1.775*10^(-3) N

part f:

magnitude of net force=sqrt(0.6169^2+1.775^2)*10^(-3) N=1.8791*10^(-3) N

part g:

angle with +ve x axis=arctan(vertical force/horizontal force)

=arctan(-1.775/0.6169)

=-70.835 degrees

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